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what happens to reflected energy ?
Roy Lewallen wrote in
: Adding a directional wattmeter to the mix raises an interesting issue. Roy, I suspect that the directional wattmeter is in a large part responsible to the strong committment to the view the Pf and Pr are separate tangible quantities, and that they can be dealt with independently. You will recall a raging argument some years ago about the possible range of rho, and insistence by some practical practitioners that rho could never be greater than unity because it would mean that reflected power was greater than forward power, obviously impossible, then citing experience with a directional wattmeter (calibrated for V/I being purely real) as evidence. Of course, telephony line designers know about this effect and do not have directional wattmeters to confuse their thinking. You example goes on to point out the absurdity of application of that kind of thinking. Miguel has asked why I keep quoting the term 'reflected power' etc, it is because it is a deceptive label that misrepresents the quantity. Roy has explained, and I have explained in the articles I referenced, the assumptions that underly the concept that Pf=Vf*If, and its limitations. There are many ham texts and articles that try to explain some concepts and deliver to the user a language and model that encourages extension to invalid application. Miguel, in a steady state scenario with a source at one end of a TEM line and a load at the other, the power (rate of flow of energy) at any point is given by the product of instantaneous voltage and current. This is a time varying quantity which can be integrated over time to obtain the average power. At some points on the line, the instantaneous power may be negative for some parts of the cycle, which indicates that over a cycle, some energy is exchanged to and fro across that point, but the average power will always be from source to load, and that quantity will decrease from source to load as accounted for by line loss elements, though in the general case, not purely exponentially as sometimes believed. Owen |
what happens to reflected energy ?
On Jun 17, 4:45*pm, Roy Lewallen wrote:
Recalling my earlier example of a 100 watt transmitter connected to a half wavelength 200 ohm transmission line and then to a 50 ohm resistive load, the "forward power" on the line was 156.25 watts and the "reverse power" or "reflected power" 56.25 watts. The subject of this topic asks where the "reflected energy" goes. Does it go back into the transmitter? ------Z01=50------+------Z02=200------load Pfor1 = 100w, Pref1 = 0w Pfor2 = 156.25w, Pref2 = 56.25w rho^2 = 0.36, rho = 0.6 That's obviously a Z0-match to 50 ohms. There are, as usual, four wavefront components: Pfor1(rho^2) = 36w, the amount of forward power on the 50 ohm line that is reflected back toward the source. Pfor1(1-rho^2) = 64w, the amount of forward power on the 50 ohm line that is transmitted through the impedance discontinuity toward the load. Pref2(rho^2) = 20.25w, the amount of reflected power on the 200 ohm line that is re-reflected back toward the load by the impedance discontinuity. Pref2(1-rho^2) = 36w, the amount of reflected power on the 200 ohm line that is transmitted through the impedance discontinuity toward the source. Pref1 = Pfor1(rho^2) + Pref2(1-rho^2) - total destructive interference Pref1 = 36w + 36w - 2*SQRT(36w*36w) = 0 Pfor2 = Pfor1(1-rho^2) + Pref2(rho^2) + constructive interference Pfor2 = 64w + 20.25w + 2*SQRT(64w*20.25w) Pfor2 = 64w + 20.25w + 72w = 156.25w Note that the 72 watts of total destructive interference energy toward the source has changed directions and joined the forward wave toward the load as constructive interference. No need to haul out a wattmeter or SWR meter. A pencil, paper, and calculator is all one needs PLUS an understanding of the rules of physics governing energy flow at a Z0-match. Actually, you didn't even need to tell me what the Pfor2 and Pref2 values are. As you can see from the calculations, they are easy to calculate even when they are unknown. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 17, 5:53*pm, K1TTT wrote:
On Jun 17, 10:31*am, Keith Dysart wrote: On Jun 14, 7:04 pm, K1TTT wrote: On Jun 14, 12:09 pm, Keith Dysart wrote: Choose an arbitrary point on the line and observe the voltage. The observed voltage will be a function of time. Let us call it V(t). At the same point on the line, observe the current and call it I(t).. if you can choose an arbitrary point, i can choose an arbitrary time... call it a VERY long time after the load is disconnected. That works for me, as long as the time is at least twice the propagation time down the line. The energy flowing (i.e. power) at this point on the line will be P(t)=V(t)I(t). If the voltage is measured in volts and the current in amps, then the power will be in joules/s or watts. We can also compute the average energy flow (power) over some interval by integrating P(t) over the interval and dividing by the length of the interval. Call this Pavg. For repetitive signals it is convenient to use one period as the interval. Note that V(t) and I(t) are arbitrary functions. The analysis applies regardless of whether V(t) is DC, a sinusoid, a step, a pulse, etc.; any arbitrary function. To provide some examples, consider the following simple setup: 1) 50 ohm, Thevenin equivalent circuit provides 100 V in to an * *open circuit. this is the case we were discussing... the others are irrelevant at this point 2) The generator is connected to a length of 50 ohm ideal * *transmission line. We will stick with ideal lines for the * *moment, because, until ideal lines are understood, there * *is no hope of understanding more complex models. 3) The ideal line is terminated in a 50 ohm resistor. snip the experiments you do not consider relevent Now let us remove the terminating resistor. The line is now open circuit. OK, now we are back to where i had objections above. Again, set the generator to DC. After one round trip time, the observations at any point on the line will be: * V(t)=100V * I(t)=0A * P(t)=0W * Pavg=0W ok, p(t)=0w, i=0a, i connect my directional wattmeter at some very long time after the load is disconnected so the transients can be ignored and i measure 0w forever. Not correct. See the equations for a directional wattmeter below. there is no power flowing in the line. I agree with this, though a directional wattmeter indicates otherwise. See below. Now it is time to introduce forward and reflected waves to the discussion. Forward an reflected waves are simply a mathematical transformation of V(t) and I(t) to another representation which can make solving problems simpler. The transformation begins with assuming that there is a forward wave defined by Vf(t)=If(t)*Z0, a reflected wave defined by Vr(t)=Ir(t)*Z0 and that at any point on the line V(t)=Vf(t)+Vr(t) and I(t)=If(t)-Ir(t). Since we are considering an ideal line, Z0 simplifies to R0. Simple algebra yields: * Vf(t)=(V(t)+R0*I(t))/2 * Vr(t)=(V(t)-R0*I(t))/2 * If(t)=Vf(t)/R0 * Ir(t)=Vr(t)/R0 The above equations are used in a directional wattmeter. Now these algebraically transformed expressions have all the capabilities of the original so they work for DC, 60 HZ, RF, sinusoids, steps, pulses, you name it. hmmm, they work for all waveforms you say... even dc?? *ok, open circuit load, z0=50, dc source of 100v with 50ohm thevenin resistance... You have made some arithmetic errors below. I have re-arranged your prose a bit to allow correction, then comment. well, maybe i missed something... lets look at the voltages: Vf(t)=(100+50*2)/2 = 100 Vr(t)=(100-50*2)/2 = 0 You made a substitution error here. On the transmission line with a constant DC voltage: * V(t)=100V * I(t)=0A substituting in to * Vf(t)=(V(t)+R0*I(t))/2 * Vr(t)=(V(t)-R0*I(t))/2 yields * Vf(t)=50V * Vr(t)=50V and * If(t) = 50/50 = 1a * Ir(t) = 50/50 = 1a Recalling that the definition of forward and reflected is * V(t)=Vf(t)+Vr(t) * I(t)=If(t)-Ir(t) we can check our arithmetic * V(t)= 50+50 = 100V * I(t)= 1-1 * = 0A Exactly as measured. If you are disagree with the results, please point out my error in the definitions, derivations or arithmetic. except i can measure If and Ir separately and can see they are both zero. * How would you measure them? The standard definition of forward and reflected waves are as I provided above. I suppose you could construct your own definitions, but such a new definition would be confusing and unlikely to have the nice properties of the standard definitions. this is where your argument about putting a resistor between two batteries falls apart... there is a length of cable with a finite time delay in the picture here. *so it is possible to actually watch the waves reflect back and forth and see what really happens. *in the case where the line is open at the far end we can measure the voltage at both ends of the line and see the constant 100v, so where is the voltage difference needed to support the forward and reverse current waves? * Well, as I suggested, the forward and reflected wave are somewhat fictitious. They are very useful as intermediate results in solving problems but, as you note above, quite problematic if one starts to assign too much reality to them. That is why they are quite analogous to the two currents in the two battery example. you can't have current without a voltage difference. * You can, though, have an arbitrary current as an intermediate result in solving a problem. With the two battery example, reduce the resistor to 0 and an infinitely large current is computed in both directions, though we know in reality that no current is actually flowing. plus i can substitute in a lossy line and measure the line heating and measure no i^2r loss that would exist if there were currents flowing in the line, so there are none. I certainly agree that there is no real current flowing, but the fictititious If and Ir are indeed present (or pick a better word if you like). try this case, which is something we do at work... take a 14ga piece of copper wire maybe 50' above ground, its characteristic impedance is probably a few hundred ohms, and put 100kv on it... how much power is it dissipating from your If and Ir heating losses?? *lets say 200 ohms and 100kv gives 100,000/200a=500a, shouldn't take long to burn up that wire should it? *and yet it never does... so why doesn't it have your circulating currents I really like this example. I thought of using something similar once to convince Cecil, but never did. He is unreedemable. But your example aptly demonstrates why If and Ir are not real, though still useful because If-Ir is the actual current flowing. Just as an aside, your comment above about requiring a voltage to cause a current to flow is not quite correct. An ideal conductor has zero resistance, so current can flow without voltage in an ideal conductor. Strangely, this can even be achieved in the real world with superconductors. So I offer another example for your consideration. The experiment is the same as above (100V, 50 ohm, ideal source and conductors) but this time short the end of the transmission line. After waiting for the line to settle, we find 0 volts everywhere and 2A flowing. Recalling the coversion equations: Vf(t)=(V(t)+R0*I(t))/2 Vr(t)=(V(t)-R0*I(t))/2 If(t)=Vf(t)/R0 Ir(t)=Vr(t)/R0 we obtain: Vf(t)=(0+100)/2 = 50V Vr(t)=(0-100)/2 = -50V If(t)=1A Ir(t)=-1A Substituting to verify V(t)=50+(-50)=0V I(t)=1-(-1)=2A Exactly as measured. I will caution again, do not assign too much reality to these forward and reflected waves; very useful for solving problems, but trouble if carried too far. ....Keith |
what happens to reflected energy ?
On Jun 17, 7:00*pm, Keith Dysart wrote:
An ideal conductor has zero resistance, so current can flow without voltage in an ideal conductor. Quoting "Fields and Waves ...", by Ramo and Whinnery: "A perfect conductor is usually understood to be a material in which there is no electric field at any frequency. Maxwell's equations ensure that there is then also no time-varying magnetic field in the perfect conductor." How does one go from zero current to a non-zero current if the magnetic field is prohibited from varying (changing) with time? -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 17, 7:00*pm, Keith Dysart wrote:
I will caution again, do not assign too much reality to these forward and reflected waves; very useful for solving problems, but trouble if carried too far. I'm afraid you are too late with that advice. Optical physicists have assigned reality to forward and reflected waves inside an interferometer and have tracked all of the energy including the energy in canceled wavefronts. http://www.teachspin.com/instruments...eriments.shtml Scroll down to, "Using Dielectric Beamsplitters to find the "missing energy" in destructive interference" It says: "... when interference is destructive at the standard output, it is constructive at the non- standard output." In RF terms, when interference is destructive at a Z0-match, it is constructive in the direction of the load. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On 17 jun, 21:30, Cecil Moore wrote:
On Jun 17, 7:00*pm, Keith Dysart wrote: An ideal conductor has zero resistance, so current can flow without voltage in an ideal conductor. Quoting "Fields and Waves ...", by Ramo and Whinnery: "A perfect conductor is usually understood to be a material in which there is no electric field at any frequency. Maxwell's equations ensure that there is then also no time-varying magnetic field in the perfect conductor." How does one go from zero current to a non-zero current if the magnetic field is prohibited from varying (changing) with time? -- 73, Cecil, w5dxp.com An ideal conductor has zero resistance, so current can flow without voltage in an ideal conductor. Inertia first Newton law applied to the free electron lttle balls perhaps? Now, if at first of experiment little balls was at rest, how do they set in movement without a force? Tell us. ...... Keith, OM, if you do not make the rope experiment, make this another simple one = get from Radio Shack a long, long lossles TL, (with vf=1, why not?), 6*10^8 meters long it is good, open or short ended (it does not matter). Connect it to your 100 W rig, key for a one second the TX full CW power, inmediately disconnect the TL and touch the connector with your fingers. Just count: tree, two, one, ¡zero!. If after "zero" you still with the connector in your fingers without blink, then reflected waves really have not too much reality... have some ointment for burns, may be the boys are right.) (sorry, do not be angry with me I am practicing translate some creole humor to english :D ) 73 Miguel LU6ETJ |
what happens to reflected energy ?
On 6/6/2010 9:22 AM, JC wrote:
Basic question (at least for me) for a very poor antenna matching : -100 w reach the antenna and 50 w are radiated. - 50 w are "reflected", what is their fate ? Are they definitely lost for radiation and just heat the line, the final.... JC It would be really really REALLY effing great if you dorks would take this to a suitable forum. This has nothing to do with antennas. And it is an unproductive infinite length, in case you bunch have missed that point. But I'm guessing that given how intelligent you all are, you already figured that out, and are ignoring it. No one is changing their position, and they never will. And yes, I have the thread set to "ignore". Which everyone should. I will NOT see your responses. tom K0TAR |
what happens to reflected energy ?
On 17 jun, 18:45, Roy Lewallen wrote:
Owen Duffy wrote: While Zo of transmission lines might not be purely real, if the sampler element is calibrated for V/I being real, then the power is given by 'forward power' - 'reflected power'. This is true even if the calibration impedance is different to the transmission line in which the measurements are made, though significant departure will impact measurement uncertainty. For example, if I insert a 50 ohm Bird 43 in a 75 ohm line at 1.8MHz, and measure Pf=150W and Pr=50W, then the power is 150-50=100W. *The insertion VSWR due to the Bird 43 is trivial in this case, so it hardly disturbs the thing being measured. Each of the power measurements is of little value alone, no inference can be made (in this case) of the actual line VSWR, but the difference of the Pf and Pr readings does give the power at that point. I discuss this in my article entitled "http://vk1od.net/blog/?p=1004" at http://vk1od.net/blog/?p=1004. Owen Adding a directional wattmeter to the mix raises an interesting issue. Recalling my earlier example of a 100 watt transmitter connected to a half wavelength 200 ohm transmission line and then to a 50 ohm resistive load, the "forward power" on the line was 156.25 watts and the "reverse power" or "reflected power" 56.25 watts. The subject of this topic asks where the "reflected energy" goes. Does it go back into the transmitter? Well, let's put a 50 ohm directional wattmeter between the transmitter and the line. It measures 100 watts forward and 0 watts reverse. So what happened to the "reverse power" on the transmission line? If it was going back into the transmitter, we've now fixed the problem just by adding the wattmeter. In fact, we could replace the wattmeter with a piece of 50 ohm transmission line of any length, as short or long as we want, and the "forward power" on that line will be 100 watts and "reverse power" zero. So we've protected the transmitter from this horrible, damaging "reverse power" just by adding a couple of inches of 50 ohm line -- or a directional wattmeter. Is this cool or what? Or we can put the wattmeter at the far end of the line and read 100 watts forward and zero watts reverse, eliminating "reverse power" at the load -- although we've lost 56.25 watts of "forward power". Or put it at the center of the line, where we'd read a whopping 451.6 watts forward and 351.6 reverse(*). These are the "forward power" and "reverse power" on the short 50 ohm transmission line (or lumped equivalent) inside the wattmeter. So we can create "forward power" and "reverse power" just by moving the wattmeter around. And most importantly, we can use it to isolate the transmitter from that bad "reverse power". What a powerful tool! I should point out that the example doesn't even mention, or need to mention, the transmitter output impedance. The entire analysis holds for any transmitter impedance, whether "dissipative", "non-dissipative", linear, or nonlinear. Wherever the "forward power" and "reverse power" come from and do to, it doesn't depend on any particular value or kind of transmitter impedance. (*) This brings up a great idea. Drop down to Radio Shack or HRO and pick up one of those circulator things that separates forward and reverse power. Put it into the middle of the line where you have 451.6 watts of "forward power" and 351.6 watts of "reverse power". Take the 351.6 watts of "reverse power", rectify it, send it to an inverter, and use it to run the transmitter -- you'll have plenty, even with poor efficiency, and you can unplug the transmitter from the mains. There should even be enough power left over to run your cooler and keep a six-pack cold. Then work DX with your 451.6 watts of forward power while you enjoy a cool one. Goodbye to electric bills! Hello to DX, free power, and a tall cool one! Roy Lewallen, W7EL- Ocultar texto de la cita - - Mostrar texto de la cita - Dear Roy: This is a brilliant piece of work, it clarify so much misconceptions (or extension of certain concepts beyond its scope). However the question of "why" persist floating somehow. Without leaving conventional approachs I think in the article cited in the "Where does it go?" thread there are some helpful and valuable hints: http://www.ittc.ku.edu/~jstiles/622/...es_package.pdf Page 25 = "The precise values of Vo+, Io+ Vo+ and Io- are therefore determined by satisfying the boundary conditions applied at each end ofthe transmission line. Page 28 = Although Vo+- and Io+- are determined by boundary conditions (i.e., what’s connected to either end of the transmission line), the ratio V+- / V+- is determined by the parameters of the transmission line only ( R, L, G, C). What it is your opinion about? 73 - Miguel Ghezzi - LU6ETJ |
what happens to reflected energy ?
lu6etj wrote:
Dear Roy: This is a brilliant piece of work, it clarify so much misconceptions (or extension of certain concepts beyond its scope). However the question of "why" persist floating somehow. Without leaving conventional approachs I think in the article cited in the "Where does it go?" thread there are some helpful and valuable hints: http://www.ittc.ku.edu/~jstiles/622/...es_package.pdf Page 25 = "The precise values of Vo+, Io+ Vo+ and Io- are therefore determined by satisfying the boundary conditions applied at each end ofthe transmission line. Page 28 = Although Vo+- and Io+- are determined by boundary conditions (i.e., what’s connected to either end of the transmission line), the ratio V+- / V+- is determined by the parameters of the transmission line only ( R, L, G, C). What it is your opinion about? 73 - Miguel Ghezzi - LU6ETJ Hello Miguel, The part of the paper you quote is very standard transmission line analysis which can be found in any textbook on the subject -- although I find this treatment to be better written than most. Nothing I have ever posted contradicts, or intends to contradict, the very well known and established theory which has been found to be correct and useful for over a hundred years. What I find silly is the idea that there are waves of average power bouncing back and forth, needing to "go" somewhere, and the specious arguments put forth in desperate attempts to explain where these elusive waves "go" and how they supposedly behave. (But interestingly, people seem much less concerned about where that extra "forward power" comes from.) It's very much like a pseudo-scientific exposition on the nature of ghosts or arguments about the effects of various astrological signs. As you've hopefully seen from my few recent postings, it's painfully simple to show that most of the conceptions about these power waves are ridiculous. But one can't change someone's religion by means of rational arguments, and it's a fool's errand to try. So my postings aren't directed to the power-wave zealots but rather to the few people who are still open to reason. I see you're one of them, and I'm glad you've giving the topic some careful thought. Roy Lewallen, W7EL |
what happens to reflected energy ?
On Jun 17, 10:57*pm, lu6etj wrote:
On 17 jun, 21:30, Cecil Moore wrote: On Jun 17, 7:00*pm, Keith Dysart wrote: An ideal conductor has zero resistance, so current can flow without voltage in an ideal conductor. Quoting "Fields and Waves ...", by Ramo and Whinnery: "A perfect conductor is usually understood to be a material in which there is no electric field at any frequency. Maxwell's equations ensure that there is then also no time-varying magnetic field in the perfect conductor." How does one go from zero current to a non-zero current if the magnetic field is prohibited from varying (changing) with time? -- 73, Cecil, w5dxp.com An ideal conductor has zero resistance, so current can flow without voltage in an ideal conductor. Inertia first Newton law applied to the free electron lttle balls perhaps? Now, if at first of experiment little balls was at rest, how do they set in movement without a force? Tell us. ..... Keith, OM, *if you do not make the rope experiment, *make this another simple one = *get from Radio Shack a long, long lossles TL, (with vf=1, why not?), 6*10^8 meters long it is good, open or short ended (it does not matter). Connect it to your 100 W rig, key for a one second the TX full CW power, inmediately disconnect the TL and touch the connector with your fingers. Just count: tree, two, one, ¡zero!. If after "zero" you still with the connector in your fingers without blink, then reflected waves really have not too much reality... have some ointment for burns, may be the boys are right.) (sorry, do not be angry with me I am practicing translate some creole humor to english :D *) 73 Miguel LU6ETJ- Hide quoted text - - Show quoted text - Good day Miguel, In the example you provide, there would be real energy flowing and it would be clearly shown by P(t)=V(t)*I(t) P(t) will be negative because the energy is flowing in the reverse direction. Computing Pf and Pr would reveal Pf(t)=0 Pr(t)=-P(t) So Pf(t)-Pr(t) = P(t) as it must. I see no conflict with anything that I have written previously. There are many examples where one can observe energy flow in a reflected wave. It just takes one counter-example to demonstrate that this is not always the case and that one should not assign *too* much reality to such waves. What this neans is that when someone asks "what happesn to reflected energy?", the first question you have to answer is "Is this a situation where the computed reflected power represents something real?" because if it does not, the original question is moot. ....Keith |
what happens to reflected energy ?
On Jun 18, 12:00*am, Keith Dysart wrote:
On Jun 17, 5:53*pm, K1TTT wrote: On Jun 17, 10:31*am, Keith Dysart wrote: On Jun 14, 7:04 pm, K1TTT wrote: On Jun 14, 12:09 pm, Keith Dysart wrote: Choose an arbitrary point on the line and observe the voltage. The observed voltage will be a function of time. Let us call it V(t). At the same point on the line, observe the current and call it I(t). if you can choose an arbitrary point, i can choose an arbitrary time... call it a VERY long time after the load is disconnected. That works for me, as long as the time is at least twice the propagation time down the line. The energy flowing (i.e. power) at this point on the line will be P(t)=V(t)I(t). If the voltage is measured in volts and the current in amps, then the power will be in joules/s or watts. We can also compute the average energy flow (power) over some interval by integrating P(t) over the interval and dividing by the length of the interval. Call this Pavg. For repetitive signals it is convenient to use one period as the interval. Note that V(t) and I(t) are arbitrary functions. The analysis applies regardless of whether V(t) is DC, a sinusoid, a step, a pulse, etc.; any arbitrary function. To provide some examples, consider the following simple setup: 1) 50 ohm, Thevenin equivalent circuit provides 100 V in to an * *open circuit. this is the case we were discussing... the others are irrelevant at this point 2) The generator is connected to a length of 50 ohm ideal * *transmission line. We will stick with ideal lines for the * *moment, because, until ideal lines are understood, there * *is no hope of understanding more complex models. 3) The ideal line is terminated in a 50 ohm resistor. snip the experiments you do not consider relevent Now let us remove the terminating resistor. The line is now open circuit. OK, now we are back to where i had objections above. Again, set the generator to DC. After one round trip time, the observations at any point on the line will be: * V(t)=100V * I(t)=0A * P(t)=0W * Pavg=0W ok, p(t)=0w, i=0a, i connect my directional wattmeter at some very long time after the load is disconnected so the transients can be ignored and i measure 0w forever. Not correct. See the equations for a directional wattmeter below. there is no power flowing in the line. I agree with this, though a directional wattmeter indicates otherwise. See below. Now it is time to introduce forward and reflected waves to the discussion. Forward an reflected waves are simply a mathematical transformation of V(t) and I(t) to another representation which can make solving problems simpler. The transformation begins with assuming that there is a forward wave defined by Vf(t)=If(t)*Z0, a reflected wave defined by Vr(t)=Ir(t)*Z0 and that at any point on the line V(t)=Vf(t)+Vr(t) and I(t)=If(t)-Ir(t). Since we are considering an ideal line, Z0 simplifies to R0. Simple algebra yields: * Vf(t)=(V(t)+R0*I(t))/2 * Vr(t)=(V(t)-R0*I(t))/2 * If(t)=Vf(t)/R0 * Ir(t)=Vr(t)/R0 The above equations are used in a directional wattmeter. Now these algebraically transformed expressions have all the capabilities of the original so they work for DC, 60 HZ, RF, sinusoids, steps, pulses, you name it. hmmm, they work for all waveforms you say... even dc?? *ok, open circuit load, z0=50, dc source of 100v with 50ohm thevenin resistance... You have made some arithmetic errors below. I have re-arranged your prose a bit to allow correction, then comment. well, maybe i missed something... lets look at the voltages: Vf(t)=(100+50*2)/2 = 100 Vr(t)=(100-50*2)/2 = 0 You made a substitution error here. On the transmission line with a constant DC voltage: * V(t)=100V * I(t)=0A substituting in to * Vf(t)=(V(t)+R0*I(t))/2 * Vr(t)=(V(t)-R0*I(t))/2 yields * Vf(t)=50V * Vr(t)=50V and * If(t) = 50/50 = 1a * Ir(t) = 50/50 = 1a Recalling that the definition of forward and reflected is * V(t)=Vf(t)+Vr(t) * I(t)=If(t)-Ir(t) we can check our arithmetic * V(t)= 50+50 = 100V * I(t)= 1-1 * = 0A Exactly as measured. If you are disagree with the results, please point out my error in the definitions, derivations or arithmetic. except i can measure If and Ir separately and can see they are both zero. * How would you measure them? you said you had a directional wattmeter, so do i. mine reads zero, what does yours read? The standard definition of forward and reflected waves are as I provided above. I suppose you could construct your own definitions, but such a new definition would be confusing and unlikely to have the nice properties of the standard definitions. this is where your argument about putting a resistor between two batteries falls apart... there is a length of cable with a finite time delay in the picture here. *so it is possible to actually watch the waves reflect back and forth and see what really happens. *in the case where the line is open at the far end we can measure the voltage at both ends of the line and see the constant 100v, so where is the voltage difference needed to support the forward and reverse current waves? * Well, as I suggested, the forward and reflected wave are somewhat fictitious. They are very useful as intermediate results in solving problems but, as you note above, quite problematic if one starts to assign too much reality to them. fictitious?? i can measure them, watch them flow from here to there, and see physical effects of them, doesn't sound very fictitious to me. weren't you the one who teaches tdr use? what are you seeing in the steps of the tdr if not for reflected waves coming back to your scope? That is why they are quite analogous to the two currents in the two battery example. no, that is why they aren't analogous. lumped components can't represent the traveling waves. you can't have current without a voltage difference. * You can, though, have an arbitrary current as an intermediate result in solving a problem. With the two battery example, reduce the resistor to 0 and an infinitely large current is computed in both directions, though we know in reality that no current is actually flowing. no it isn't. in the limit at R-0 0V/R still looks like 0 to me. plus i can substitute in a lossy line and measure the line heating and measure no i^2r loss that would exist if there were currents flowing in the line, so there are none. I certainly agree that there is no real current flowing, but the fictititious If and Ir are indeed present (or pick a better word if you like). now you say they are fictitious, yet you have formulas for them and can measure them... until you come to your senses and admit that reflections are real there isn't much we can do for you so the rest of your post is based on an incorrect premise. snip |
what happens to reflected energy ?
On 6/17/2010 9:59 PM, tom wrote:
On 6/6/2010 9:22 AM, JC wrote: Basic question (at least for me) for a very poor antenna matching : -100 w reach the antenna and 50 w are radiated. - 50 w are "reflected", what is their fate ? Are they definitely lost for radiation and just heat the line, the final.... JC It would be really really REALLY effing great if you dorks would take this to a suitable forum. This has nothing to do with antennas. And it is an unproductive infinite length, in case you bunch have missed that point. But I'm guessing that given how intelligent you all are, you already figured that out, and are ignoring it. No one is changing their position, and they never will. And yes, I have the thread set to "ignore". Which everyone should. I will NOT see your responses. tom K0TAR Actually, I'm following and enjoying the conversation. Since you set your computer to ignore responses, then you should have no objection if they continue. It makes more sense to ignore subjects which don't interest you than to complain about them. John KD5YI |
what happens to reflected energy ?
On Jun 17, 9:59*pm, tom wrote:
It would be really really REALLY effing great if you dorks would take this to a suitable forum. *This has nothing to do with antennas. Is there a rec.radio.amateur.feedlines group that I have missed? -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 18, 3:06*am, Roy Lewallen wrote:
Nothing I have ever posted contradicts, or intends to contradict, the very well known and established theory which has been found to be correct and useful for over a hundred years. On the contrary, what you have said indeed does contradict 100 years of optical physics. All you have to do to alleviate your ignorance is to grok that superposition (wave-cancellation/interference) can and does redistribute energy even when there are no reflections present. The ONLY time that a forward wave and a reflected wave don't interfere with each other is when they are 90 degrees different in phase. What I find silly is the idea that there are waves of average power bouncing back and forth, needing to "go" somewhere, and the specious arguments put forth in desperate attempts to explain where these elusive waves "go" and how they supposedly behave. Have you never stood in a hall of mirrors where the reflections seem to go on out to infinity on both sides? You would have us believe that those reflections are not bouncing back and forth, mirror to mirror, and that those reflections that you can see with your own eyes have zero energy??? Roy, you are just spouting guru-type metaphysics, and should be ashamed of yourself for trying to spread your religion to the unwashed masses. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 18, 6:00*am, Keith Dysart wrote:
"Is this a situation where the computed reflected power represents something real?" because if it does not, the original question is moot. It certainly depends upon your definition of "power" and that definition is different between pure physics and RF engineering. One of the accepted definitions of "power" in "The IEEE Dictionary" contradicts the definition of "power" given in my college physics book. "The IEEE Dictionary" says that flowing energy passing a fixed measurement point is power, by definition. If there is anything at all that can be measured, it must necessarily contain energy. If the EM energy is moving, it is power, by IEEE definition. If you disagree, take it up with The IEEE. So the actual question that needs to be answered is: Does the electromagnetic reflected wave contain energy traveling at the speed of light in the medium? The answer is yes, being photonic in nature, an EM wave must necessarily contain an ExH power density and, consisting of photons, must necessarily be traveling at the speed of light in the medium. The EM wave will continue to travel in the direction of energy flow at the speed of light in the medium until it encounters an impedance discontinuity which causes a reflection and/or interference. There is really no difference in a forward wave and a reflected wave except for the direction of travel. They both contain an associated ExH power density. One can set up identical signal generators at each end of a transmission line to emulate forward and reflected waves. Which generator yields forward waves and which generator yields reflected waves? It doesn't really matter because direction is only a convention. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On 18 jun, 12:26, Cecil Moore wrote:
On Jun 18, 6:00*am, Keith Dysart wrote: "Is this a situation where the computed reflected power represents something real?" because if it does not, the original question is moot. It certainly depends upon your definition of "power" and that definition is different between pure physics and RF engineering. One of the accepted definitions of "power" in "The IEEE Dictionary" contradicts the definition of "power" given in my college physics book. "The IEEE Dictionary" says that flowing energy passing a fixed measurement point is power, by definition. If there is anything at all that can be measured, it must necessarily contain energy. If the EM energy is moving, it is power, by IEEE definition. If you disagree, take it up with The IEEE. So the actual question that needs to be answered is: Does the electromagnetic reflected wave contain energy traveling at the speed of light in the medium? The answer is yes, being photonic in nature, an EM wave must necessarily contain an ExH power density and, consisting of photons, must necessarily be traveling at the speed of light in the medium. The EM wave will continue to travel in the direction of energy flow at the speed of light in the medium until it encounters an impedance discontinuity which causes a reflection and/or interference. There is really no difference in a forward wave and a reflected wave except for the direction of travel. They both contain an associated ExH power density. One can set up identical signal generators at each end of a transmission line to emulate forward and reflected waves. Which generator yields forward waves and which generator yields reflected waves? It doesn't really matter because direction is only a convention. -- 73, Cecil, w5dxp.com Hello friends. good day for you. I am glad that I have not bothered with my joke, Keith! :) I was to do some comments to your and Roy posts, but I prefer to ask questions rather than giving personal opinions :) ..... Please Cecil, friend, do not mention photons and optics (do not bother me, don't worry), I believe you better play your game without such things. I know is like fighting with one arm tied, but I think you can, make an effort :) Please Owen explain to me what is "negative power". In the other thread you said = "The notion that reflected power is simply and always absorbed in the real source resistance is quite wrong. Sure you can build special cases where that might happen, but there is more to it. Thinking of the reflected wave as 'reflected power' leads to some of the misconception. Then I asked; "remember me what reflected power definition are you using here and expand the sentence idea". I did not ask for the quoting, I ask (again) for your concept of Reflected Power. I do not sure if you accept (or not) the very notion of reflected power as legitimate. The sentence it is not clear enough to me. Please, tell us if you accept the idea of two very directive electromagnetic waves from different sources flowing in opposite directions as carring independent energy (with different frequencies at first) and information, and the posibbility that energy be dissipated and/or transmitted by de opposite system and the information being recovered in both ends. If your answer is yes, then tell us if you conceive such similar system using a same wave guide to simultaneously vinculate both system. 73 - Miguel - LU6ETJ PS: Roy, friend, take it ease (is OK, or polite say "take it easy?), references to the article were only to complement (bring close?) some useful notions related, mentioning a possible common reference (my english books are few, old and possibly useless as reference nowadays) |
what happens to reflected energy ?
On Jun 18, 1:01*pm, lu6etj wrote:
Please Cecil, friend, do not mention photons and optics (do not bother me, don't worry), ... What do I do about people who assert that RF waves possess the ability to violate the known laws of EM wave physics? It doesn't matter what the frequency of an EM wave is, it must obey the laws of physics, two of which a 1. It cannot exist without a Poynting vector ExH power density. 2. It must necessarily move at the speed of light in the medium. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 18, 7:04*am, K1TTT wrote:
On Jun 18, 12:00*am, Keith Dysart wrote: except i can measure If and Ir separately and can see they are both zero. * How would you measure them? you said you had a directional wattmeter, so do i. *mine reads zero, what does yours read? That would most likely be because your wattmeter is AC coupled. Working with DC, you need a DC-coupled meter, in which case it would read as I predicted. A highly instructive exercise is to examine the schematic for your directional wattmeter and work out how it implements the expressions Vf(t)=(V(t)+R0*I(t))/2 Vr(t)=(V(t)-R0*I(t))/2 Pf=avg(Vf(t)*Vf(t)/R0) Pr=avg(Vr(t)*Vr(t)/R0) or some equivalent variant using If and Ir. You will find a voltage sampler, a current sampler, some scaling, and an addition (or subtraction). If your meter is analog, the squaring function is usually implemented in the non-linear meter scale. And the averaging function might be peak-reading or rely on meter inertia. You can, though, have an arbitrary current as an intermediate result in solving a problem. With the two battery example, reduce the resistor to 0 and an infinitely large current is computed in both directions, though we know in reality that no current is actually flowing. no it isn't. *in the limit at R-0 *0V/R still looks like 0 to me. When computing the intermediate results using superposition, you have the full battery voltage in to 0 ohms. First the battery on the left, then the one on the right, and then you add the two currents to get the total current. 0 is always the total, but the intermediate results can go to infinity if you connect the two batteries directly in parallel. I certainly agree that there is no real current flowing, but the fictititious If and Ir are indeed present (or pick a better word if you like). now you say they are fictitious, yet you have formulas for them and can measure them... until you come to your senses and admit that reflections are real there isn't much we can do for you so the rest of your post is based on an incorrect premise. Then what are the equations for computing forward and reflected waves and 'power'? That is where I started. Is that the premise you consider to be incorrect? Do follow through the arithmetic in the previously provided examples and see if you can find any faults. Point me to any errors and I will gladly reconsider my position. If not, perhaps it is time for you to start questioning your assumptions. ....Keith |
what happens to reflected energy ?
On Jun 18, 11:26 pm, Keith Dysart wrote:
On Jun 18, 7:04 am, K1TTT wrote: On Jun 18, 12:00 am, Keith Dysart wrote: except i can measure If and Ir separately and can see they are both zero. How would you measure them? you said you had a directional wattmeter, so do i. mine reads zero, what does yours read? That would most likely be because your wattmeter is AC coupled. Working with DC, you need a DC-coupled meter, in which case it would read as I predicted. mine is dc coupled... still nothing. please provide the schematic or model number for the one you use to get such a reading. A highly instructive exercise is to examine the schematic for your directional wattmeter and work out how it implements the expressions Vf(t)=(V(t)+R0*I(t))/2 Vr(t)=(V(t)-R0*I(t))/2 Pf=avg(Vf(t)*Vf(t)/R0) Pr=avg(Vr(t)*Vr(t)/R0) or some equivalent variant using If and Ir. You will find a voltage sampler, a current sampler, some scaling, and an addition (or subtraction). If your meter is analog, the squaring function is usually implemented in the non-linear meter scale. And the averaging function might be peak-reading or rely on meter inertia. You can, though, have an arbitrary current as an intermediate result in solving a problem. With the two battery example, reduce the resistor to 0 and an infinitely large current is computed in both directions, though we know in reality that no current is actually flowing. no it isn't. in the limit at R-0 0V/R still looks like 0 to me. When computing the intermediate results using superposition, you have the full battery voltage in to 0 ohms. First the battery on the left, then the one on the right, and then you add the two currents to get the total current. 0 is always the total, but the intermediate results can go to infinity if you connect the two batteries directly in parallel. there is no battery on the right, unless you have changed the case again. i thought we were doing the open circuit coax with the dc source on one end. but even if you have the case above and connect another 100v battery on the open end of the line there will be no current flowing back as there is no voltage difference to drive it. I certainly agree that there is no real current flowing, but the fictititious If and Ir are indeed present (or pick a better word if you like). now you say they are fictitious, yet you have formulas for them and can measure them... until you come to your senses and admit that reflections are real there isn't much we can do for you so the rest of your post is based on an incorrect premise. Then what are the equations for computing forward and reflected waves and 'power'? That is where I started. Is that the premise you consider to be incorrect? yep, your 'waves' don't include the proper term to make them propagating waves that satisfy maxwell's equations... so they can't be real... so your initial premise is incorrect. Do follow through the arithmetic in the previously provided examples and see if you can find any faults. Point me to any errors and I will gladly reconsider my position. If not, perhaps it is time for you to start questioning your assumptions. ...Keith my assumptions are perfectly valid... it is yours that are flawed from the very beginning. |
what happens to reflected energy ?
On Jun 17, 9:12*am, Cecil Moore wrote:
On Jun 17, 5:38*am, Keith Dysart wrote: You are an RF dude and there is nothing that can possibly be learned from simple DC circuits. We studied DC circuits in EE101. Later we were forced to expand our knowledge into distributed networks because the DC circuit model failed at RF frequencies. One does, however, usually require that the more sophisticated model continue to provide the correct answer for the simpler problems that could be solved in another way. And it does here as well, though the answers make some uncomfortable. Still, I observe that you have discovered no errors in the exposition. I have not wasted my time time trying to discover any errors. In a nutshell, what new knowledge have you presented? Well, at least you continue to read the posts, so perhaps there is some slim hope, but I am not holding my breath. As for me, I have learned a lot from your postings since I first encountered them circa 1994. Back then, I was not so knowledgeable about transmission lines and I was trying to decide who was right. You posted well thought out, convincing arguments and I found myself switching from side to side. Reflected waves heat finals, no they don't, yes they do, ... but eventually I learned. Then I continued to read your posts, visited your web site and studied your arguments. The challenge was to discover where you had gone wrong. Occasionally I would think up a suitable counter example and offer it to you, but, even then, you refused to look at examples that might threaten your beliefs. So I no longer post with any expectation that you might be persuaded. Mostly now I post to ensure that a view other than yours is voiced to help prevent those who are not yet sure from being sucked in to your vortex. Anyhow, I suggest you read my examples and find the flaws, for that is the surest way to convince me that I am wrong. And don't just post other examples that support your position, for it only takes one counter-example to disprove a hypothesis, no matter how many examples are in agreement. And don't reject simple DC examples because they lead to uncomfortable answers for it is by examining the examples that do not support your hypotheses that you will learn, not by sticking just with the ones that do. ....Keith |
what happens to reflected energy ?
On Jun 18, 7:55*pm, K1TTT wrote:
On Jun 18, 11:26 pm, Keith Dysart wrote: On Jun 18, 7:04 am, K1TTT wrote: On Jun 18, 12:00 am, Keith Dysart wrote: except i can measure If and Ir separately and can see they are both zero. How would you measure them? you said you had a directional wattmeter, so do i. *mine reads zero, what does yours read? That would most likely be because your wattmeter is AC coupled. Working with DC, you need a DC-coupled meter, in which case it would read as I predicted. mine is dc coupled... still nothing. *please provide the schematic or model number for the one you use to get such a reading. Oh my. Could you kindly provide the schematic or a reference? Before posting I spent an hour searching my library and the web for the circuit for a dc-coupled directional wattmeter and was unsuccessful. Nothing prevents one from being constructed. That one is not readily available should not prevent you from using the equations that they implement and then predicting the DC behaviour of the line. A highly instructive exercise is to examine the schematic for your directional wattmeter and work out how it implements the expressions * Vf(t)=(V(t)+R0*I(t))/2 * Vr(t)=(V(t)-R0*I(t))/2 * Pf=avg(Vf(t)*Vf(t)/R0) * Pr=avg(Vr(t)*Vr(t)/R0) or some equivalent variant using If and Ir. You will find a voltage sampler, a current sampler, some scaling, and an addition (or subtraction). If your meter is analog, the squaring function is usually implemented in the non-linear meter scale. And the averaging function might be peak-reading or rely on meter inertia. You can, though, have an arbitrary current as an intermediate result in solving a problem. With the two battery example, reduce the resistor to 0 and an infinitely large current is computed in both directions, though we know in reality that no current is actually flowing. no it isn't. *in the limit at R-0 *0V/R still looks like 0 to me.. When computing the intermediate results using superposition, you have the full battery voltage in to 0 ohms. First the battery on the left, then the one on the right, and then you add the two currents to get the total current. 0 is always the total, but the intermediate results can go to infinity if you connect the two batteries directly in parallel. there is no battery on the right, unless you have changed the case again. *i thought we were doing the open circuit coax with the dc source on one end. No. I was referring to the simple DC circuit with two batteries and a resistor. I was pretty sure that was your referent when you used "R-0". I understand that you have not found any flaws in my exposition, except that you are still uncomfortable with the results. It is possible to overcome this. And I re-iterate the value of studying your directional wattmeter's schematic, especially if it is DC-coupled. ....Keith |
what happens to reflected energy ?
On Jun 17, 8:38*pm, Cecil Moore wrote:
On Jun 17, 7:00*pm, Keith Dysart wrote: I will caution again, do not assign too much reality to these forward and reflected waves; very useful for solving problems, but trouble if carried too far. I'm afraid you are too late with that advice. Optical physicists have assigned reality to forward and reflected waves inside an interferometer and have tracked all of the energy including the energy in canceled wavefronts. http://www.teachspin.com/instruments...eriments.shtml Scroll down to, "Using Dielectric Beamsplitters to find the "missing energy" in destructive interference" It says: "... when interference is destructive at the standard output, it is constructive at the non- standard output." In RF terms, when interference is destructive at a Z0-match, it is constructive in the direction of the load. -- 73, Cecil, w5dxp.com You have fallen in to the same trap that you did with circulators. When you introduce a circulator in to the transmission line, you change the experiment and you get certain results. The circulator provides some numerology that appears to support the hypothesis of 'reflected energy' being real. The same with looking for "missing energy" (note they even quote it) with a beamsplitter. The beamsplitter reveals what is happening at the beamsplitter, not what is happening at points between the two beamsplitters. Just like a circulator. ....Keith |
what happens to reflected energy ?
On Jun 19, 12:22*am, Keith Dysart wrote:
On Jun 17, 8:38*pm, Cecil Moore wrote: On Jun 17, 7:00*pm, Keith Dysart wrote: I will caution again, do not assign too much reality to these forward and reflected waves; very useful for solving problems, but trouble if carried too far. I'm afraid you are too late with that advice. Optical physicists have assigned reality to forward and reflected waves inside an interferometer and have tracked all of the energy including the energy in canceled wavefronts. http://www.teachspin.com/instruments...eriments.shtml Scroll down to, "Using Dielectric Beamsplitters to find the "missing energy" in destructive interference" It says: "... when interference is destructive at the standard output, it is constructive at the non- standard output." In RF terms, when interference is destructive at a Z0-match, it is constructive in the direction of the load. -- 73, Cecil, w5dxp.com You have fallen in to the same trap that you did with circulators. When you introduce a circulator in to the transmission line, you change the experiment and you get certain results. The circulator provides some numerology that appears to support the hypothesis of 'reflected energy' being real. The same with looking for "missing energy" (note they even quote it) with a beamsplitter. The beamsplitter reveals what is happening at the beamsplitter, not what is happening at points between the two beamsplitters. Just like a circulator. ...Keith- Hide quoted text - - Show quoted text - he is even more hopeless than most, first he determines there are forward and reflected waves when a dc source has charged the line, then he rejects circulators and beamsplitters that obviously work because of the waves. so dc waves exist, but rf and optical waves don't. thats a unique viewpoint. |
what happens to reflected energy ?
On Jun 18, 8:36*pm, K1TTT wrote:
On Jun 19, 12:22*am, Keith Dysart wrote: On Jun 17, 8:38*pm, Cecil Moore wrote: On Jun 17, 7:00*pm, Keith Dysart wrote: I will caution again, do not assign too much reality to these forward and reflected waves; very useful for solving problems, but trouble if carried too far. I'm afraid you are too late with that advice. Optical physicists have assigned reality to forward and reflected waves inside an interferometer and have tracked all of the energy including the energy in canceled wavefronts. http://www.teachspin.com/instruments...eriments.shtml Scroll down to, "Using Dielectric Beamsplitters to find the "missing energy" in destructive interference" It says: "... when interference is destructive at the standard output, it is constructive at the non- standard output." In RF terms, when interference is destructive at a Z0-match, it is constructive in the direction of the load. -- 73, Cecil, w5dxp.com You have fallen in to the same trap that you did with circulators. When you introduce a circulator in to the transmission line, you change the experiment and you get certain results. The circulator provides some numerology that appears to support the hypothesis of 'reflected energy' being real. The same with looking for "missing energy" (note they even quote it) with a beamsplitter. The beamsplitter reveals what is happening at the beamsplitter, not what is happening at points between the two beamsplitters. Just like a circulator. ...Keith- Hide quoted text - - Show quoted text - he is even more hopeless than most, first he determines there are forward and reflected waves when a dc source has charged the line, then he rejects circulators and beamsplitters that obviously work because of the waves. *so dc waves exist, but rf and optical waves don't. *thats a unique viewpoint.- Hide quoted text - - Show quoted text - You might consider the following experiment: - a sinusoidal generator that will produce 50W in to 50 ohms - attach one half wavelength of transmission line - leave the far end of the transmission line open After the line settles, a directional "wattmeter" anywhere on the line will indicate 50W forward and 50W reflected. The generator will not be putting any energy in to the line since the line input appears as an open circuit at the generator output. Insert a circulator (with the reflected port terminated in 50 ohms) between the generator and the line. The circulator termination will be dissipating 50W and the generator will now be delivering 50W in to the circulator. 1. Please explain how inserting the circulator did not change the circuit conditions. The generator went from delivering 0W to delivering 50W. 2. Where do you think the 50W being dissipated in the circulator termination resistor is coming from? The line? Or the generator (which is now outputting 50W)? Just for fun, you may be interested in a DC coupled circulator that works all the way down to 0 Hz: http://www.techlib.com/files/RFDesign3.pdf Answer question 2 for this circulator design. ....Keith |
what happens to reflected energy ?
On Jun 18, 7:00*pm, Keith Dysart wrote:
Reflected waves heat finals, no they don't, yes they do, ... but eventually I learned. Hopefully, you learned that rail arguments are rarely valid and the truth usually lies somewhere in between. Sometimes reflected waves heat finals and sometimes they don't. It all depends on what kind and how much interference exists between the forward wave and reflected wave at the source impedance and (apparently) how much of that impedance is dissipative or non-dissipative. Roy's food-for-thought article doesn't take interference into account at all and therefore arrives at magic conclusions divorced from reality. Anyhow, I suggest you read my examples and find the flaws, for that is the surest way to convince me that I am wrong. I have pointed out many of the flaws in your examples. The way you parse the energy violates the laws of physics. Maxwell's equations only work on a function that contains classic EM traveling waves. Standing waves and DC steady-state examples don't meet the necessary boundary conditions. In fact, it can be proved that EM waves do not existent during DC steady-state. The fact that an instrument designed to detect EM waves malfunctions during DC steady-state is not unexpected. Even an SWR meter calibrated for a different Z0 than the one being used will indicate false results. You really need to learn to recognize when your experiment and your math model are contradictory, when your concepts violate the laws of physics, and when you have chosen the wrong measuring equipment. You remind me of myself when I was 14 years old and tried to measure the screen voltage on a vacuum tube using a Simpson multimeter. It wasn't anywhere near the voltage specified in the manual so I wasted my money by buying a new tube the next time my family made a trip to Houston. You would report those same measurement results as proof of a strange, magical happening. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 18, 7:22*pm, Keith Dysart wrote:
The beamsplitter reveals what is happening at the beamsplitter, not what is happening at points between the two beamsplitters. Oh yeah, according to you, the light beams are changing directions at every voltage and current node. Sorry, I do not accept your metaphysics, especially when they violate the accepted laws of physics. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 20, 8:03*pm, Keith Dysart wrote:
1. Please explain how inserting the circulator did not change * *the circuit conditions. The generator went from delivering * *0W to delivering 50W. The generator went from experiencing total destructive interference to experiencing zero interference. Of course, the entire purpose of inserting the circulator is to cause the generator to see 50 ohms as a load impedance. 2. Where do you think the 50W being dissipated in the circulator * *termination resistor is coming from? The line? Or the * *generator (which is now outputting 50W)? TV ghosting experiments will verify that the signal being dissipated in the circulator load resistor has made a round trip from the generator to the end of the N(1/2WL) stub and back to the circulator. Please don't insult our intelligence by arguing that is not a steady- state condition. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 21, 1:03*am, Keith Dysart wrote:
On Jun 18, 8:36*pm, K1TTT wrote: On Jun 19, 12:22*am, Keith Dysart wrote: On Jun 17, 8:38*pm, Cecil Moore wrote: On Jun 17, 7:00*pm, Keith Dysart wrote: I will caution again, do not assign too much reality to these forward and reflected waves; very useful for solving problems, but trouble if carried too far. I'm afraid you are too late with that advice. Optical physicists have assigned reality to forward and reflected waves inside an interferometer and have tracked all of the energy including the energy in canceled wavefronts. http://www.teachspin.com/instruments...eriments.shtml Scroll down to, "Using Dielectric Beamsplitters to find the "missing energy" in destructive interference" It says: "... when interference is destructive at the standard output, it is constructive at the non- standard output." In RF terms, when interference is destructive at a Z0-match, it is constructive in the direction of the load. -- 73, Cecil, w5dxp.com You have fallen in to the same trap that you did with circulators. When you introduce a circulator in to the transmission line, you change the experiment and you get certain results. The circulator provides some numerology that appears to support the hypothesis of 'reflected energy' being real. The same with looking for "missing energy" (note they even quote it) with a beamsplitter. The beamsplitter reveals what is happening at the beamsplitter, not what is happening at points between the two beamsplitters. Just like a circulator. ...Keith- Hide quoted text - - Show quoted text - he is even more hopeless than most, first he determines there are forward and reflected waves when a dc source has charged the line, then he rejects circulators and beamsplitters that obviously work because of the waves. *so dc waves exist, but rf and optical waves don't. *thats a unique viewpoint.- Hide quoted text - - Show quoted text - You might consider the following experiment: - a sinusoidal generator that will produce 50W in to 50 ohms - attach one half wavelength of transmission line - leave the far end of the transmission line open After the line settles, a directional "wattmeter" anywhere on the line will indicate 50W forward and 50W reflected. The generator will not be putting any energy in to the line since the line input appears as an open circuit at the generator output. Insert a circulator (with the reflected port terminated in 50 ohms) between the generator and the line. The circulator termination will be dissipating 50W and the generator will now be delivering 50W in to the circulator. 1. Please explain how inserting the circulator did not change * *the circuit conditions. The generator went from delivering * *0W to delivering 50W. 2. Where do you think the 50W being dissipated in the circulator * *termination resistor is coming from? The line? Or the * *generator (which is now outputting 50W)? well, its coming from the generator via the far end of the line of course. Just for fun, you may be interested in a DC coupled circulator that works all the way down to 0 Hz:http://www.techlib.com/files/RFDesign3.pdf Answer question 2 for this circulator design. ...Keith the result is no current, but 50 v, as would be expected of course. |
what happens to reflected energy ?
On Jun 21, 9:04*am, Cecil Moore wrote:
On Jun 20, 8:03*pm, Keith Dysart wrote: 1. Please explain how inserting the circulator did not change * *the circuit conditions. The generator went from delivering * *0W to delivering 50W. The generator went from experiencing total destructive interference to experiencing zero interference. Of course, the entire purpose of inserting the circulator is to cause the generator to see 50 ohms as a load impedance. Along with the other use for circulators: To present an impedance match to the reflected wave so that it will not be re-reflected. 2. Where do you think the 50W being dissipated in the circulator * *termination resistor is coming from? The line? Or the * *generator (which is now outputting 50W)? TV ghosting experiments will verify that the signal being dissipated in the circulator load resistor has made a round trip from the generator to the end of the N(1/2WL) stub and back to the circulator. It is not clear why you think this verifies something. Enhancing the details slightly of the generator in the original example: The generator is constructed using the Thevenin model of a source followed by a 50ohm resistor. With the circulator, there is no re-reflection at the driven end of the line, 50W is dissipated in the circulator resistor and 50W is dissipated in the source resistor. Without the circulator, there is no re-reflection and nothing is dissipated in the source resistor (and there is no circulator resistor). The dissipation seems to correlate much more strongly with the circuit design than it does with the magnitude of the "reflected power". Please don't insult our intelligence by arguing that is not a steady- state condition. Well you are right, it is not steady-state when there is modulation, but modulation does not affect the examples I have provided. ....Keith PS: For further understanding, substitute a Norton style generator, then do it again for both kinds of generators but with the end of the line shorted yielding more examples where the "reflected power" does not correlate with the dissipation. |
what happens to reflected energy ?
On Jun 21, 5:40 pm, K1TTT wrote:
On Jun 21, 1:03 am, Keith Dysart wrote: You might consider the following experiment: - a sinusoidal generator that will produce 50W in to 50 ohms - attach one half wavelength of transmission line - leave the far end of the transmission line open After the line settles, a directional "wattmeter" anywhere on the line will indicate 50W forward and 50W reflected. The generator will not be putting any energy in to the line since the line input appears as an open circuit at the generator output. Insert a circulator (with the reflected port terminated in 50 ohms) between the generator and the line. The circulator termination will be dissipating 50W and the generator will now be delivering 50W in to the circulator. 1. Please explain how inserting the circulator did not change the circuit conditions. The generator went from delivering 0W to delivering 50W. 2. Where do you think the 50W being dissipated in the circulator termination resistor is coming from? The line? Or the generator (which is now outputting 50W)? well, its coming from the generator via the far end of the line of course. That is the classic answer from the “reflected power” model, but consider the following... We construct two experiments similar to the above. The first one: - a sinusoidal generator that will produce 50W in to 50 ohms - attach one half wavelength of transmission line - leave the far end of the transmission line open - the generator is constructed in the Thevenin style with a voltage source and 50 ohm resistor The second experiment: - same as above except there is a circulator between the generator and the line and the circulator is terminated with 50 ohms Examine experiment 1. After the line settles, a directional wattmeter indicates 50W “forward power” and 50W “reflected power”. The current in the voltage source and source resistor is 0 so no energy is dissipated in the source resistor and the voltage source is delivering no energy to the system. The “reflected power” is not being dissipated in the source resistor or voltage source so where is it going. Cecil will offer some explanation where the “reflected energy” is “redistributed” in the other direction as the “forward energy”, which “must” be happening since it is not dissipated in the source and it can not be reflected because there is no impedance discontinuity. Examine experiment 2. After the line settles, a directional wattmeter indicates 50W “forward power” and 50W “reflected power”. But with a circulator, the voltage source is delivering 100W, the source resistor is dissipating 50W and the circulator resistor is also dissipating 50W. You offer that the 50W being dissipated in the circulator is the 50W of “reflected power” coming from the line. Presumably then, the 50W of “forward power” is coming from the generator. This agrees numerically since the voltage source is providing 100W and the source resistor is dissipating 50W. In both experiments, conditions on the line are identical. There is 50W of “forward and reflected power” indicated. There is no energy flowing past the end of the line since it is open circuited. Because the line is one-half wavelength long, the line presents an open circuit to the generator (exp 1) or circulator (exp 2). In both experiments, the line is presented with a 50 ohm source impedance, either from the generator or from the circulator. What puzzles me then, is how the “reflected power” knows that in experiment 1, it should stay out of the generator so that it is not dissipated in the source resistor but in experiment 2, it should enter the circulator so that it can be dissipated in the circulator load resistor. Can you explain how the “reflected power” “knows”? ....Keith |
what happens to reflected energy ?
On Jun 21, 7:25*pm, Keith Dysart wrote:
The dissipation seems to correlate much more strongly with the circuit design than it does with the magnitude of the "reflected power". Given a 50 ohm source resistor resulting in zero re-reflections, *the dissipation correlates perfectly with the level and type of interference*. I am amazed at the level of ignorance concerning interference and the conservation of energy principle. I studied interference in detail in my 1950's college physics courses including what happens to the energy during interference events. Although Dr. Best didn't realize it, in his QEX Nov/Dec 2001 article on transmission lines, he presented the equations governing energy redistribution associated with interference. What didn't help was his equation of the form: Ptot = 75w + 8.33w = 133.33w which is obviously false. What he left out was the constructive interference term of: 2*SQRT(75w*8.33w) = 50w Given two coherent waves, each associated with a Poynting Vector, when those two waves are superposed, interference will result. I'm sure everyone has seen interference rings and pictures of interference rings where the energy in the interference rings has obviously been redistributed away from a homogeneous power density. The same thing happens in a transmission line and/or at a source, just at longer wavelengths where we cannot "see" them with our eyes. Constructive interference requires "extra" energy. Superpose two 50 watt coherent waves in phase and the result is a 200 watt wave. Destructive interference has energy left over. Superpose two 50 watt coherent waves 180 degrees out of phase and the result is a zero watt wave. At a source, the source is capable of supplying extra power or throttling back on its power output depending on the level and kind of interference. If one simply takes the time to understand interference, one knows where the energy components go even when there is zero re- reflection. Given a single traveling wave in a lossless transmission line with a purely resistive characteristic impedance of Z0, the power (joules/ second) measured at a point on the line is P=V^2/Z0=I^2*Z0=V*I. This is classic transmission line math no matter which direction the wave is traveling. Superposing two coherent traveling waves cannot create or destroy the energy pre-existing in the two waves, but it can (and often does) redistribute the energy according to the conservation of energy principle. If the total energy in the two superposed waves is different from the sum of the two energy components before superposition, then interference exists and some interference energy has been redistributed in the system. That some people are ignorant of where the constructive interference energy came from or where the destructive interference energy went is not a good reason to deny its existence (which violates the accepted laws of physics). The thing that makes an interference event so easy to analyze in a transmission line is the fact that the transmission line is essentially one-dimensional. At an impedance discontinuity in a transmission line (away from any active source) if interference exists, the destructive interference in one direction must equal the constructive interference in the other direction. Any other outcome would violate the conservation of energy principle and that is a common occurence on this newsgroup. In my earlier example, the two results of *(1) reflection* are traveling waves containing ExH power densities: Pfor1(rho^2) toward the source and Pref2(rho^2) toward the load The complimentary transmissions a Pfor1(1-rho^2) toward the load and Pref2(1-rho^2) toward the source Those two reflections *(2) superpose* with the two complimentary transmissions in each same direction according to the power density equation: 0 = Pref1 = Pfor1(rho^2) + Pref2(1-rho^2) minus total destructive interference toward the source Pfor2 = Pfor1(1-rho^2) + Pref2(rho^2) plus constructive interference toward the load When one assumes that reflection is the only phenomenon that can redistribute wave energy, one is ignoring the superposition process which is also known to be perfectly capable of redistributing wave energy. In Roy's food-for-thought example, he designed it so reflections would not exist at the source. That leaves superposition as the phenomenon that is accomplishing the obvious redistribution of energy. It certainly does NOT mean that the laws of physics have been suspended due to the ignorance of the writer. Here's a simple exercise in phasor superposition. Given two 50w (joules/sec) coherent EM waves traveling in the same direction in Z0=50 ohm coax: 1. What is the magnitude of the two voltages? ________ 2. If one superposes those two voltages in phase in the Z0=50 ohm transmission line, what is the resulting power (joules/sec) in the total wave? ___________ 3. If one superposes those two voltages 180 degrees out of phase in the Z0=50 ohm transmission line, what is the resulting power (joules/ sec) in the total wave? ____________ In #2, if there is no source around, where does the "extra" wave energy come from? In #3, if there is no source or sink around, where does the "excess" wave energy go? -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 22, 5:49*am, Keith Dysart wrote:
What puzzles me then, is how the “reflected power” knows that in experiment 1, it should stay out of the generator so that it is not dissipated in the source resistor but in experiment 2, it should enter the circulator so that it can be dissipated in the circulator load resistor. Can you explain how the “reflected power” “knows”? Reflected "power" doesn't know anything. Reflected voltages and currents simply respond to the known laws of physics involving wave reflection and wave superposition, and obey the conservation of energy principle. In 1, it doesn't stay out of the generator. It is redistributed from the generator back toward the load by superposition associated with *destructive interference) which happens when two superposed coherent waves are between 90 deg and 180 deg out of phase. The entire experiment is set up in a perfect 50 ohm environment so power = V^2/50. What you guys are missing is that step which carries the energy along with the voltage, i.e. the voltage (and current) require a certain energy level which is being ignored. In 2, the source wave never encounters the reflected wave so there is no superposition or interference at the source resistor. You can add a couple of more configurations. Number 3 could be when the interference between the source wave and the reflected wave occurs and they are less than 90 deg out of phase. More power than the average reflected power plus average load power will be dissipated in the source resistor because of superposition associated with *constructive interference*. Number 4 could be the special case when the source wave and the reflected wave are 90 degrees out of phase. This is the condition of zero interference (neither destructive nor constructive) and 100% of the average reflected power is dissipated in the source resistor. This is the easiest case to understand because there are no reflections and no interference. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On 22 jun, 11:45, Cecil Moore wrote:
On Jun 22, 5:49*am, Keith Dysart wrote: What puzzles me then, is how the “reflected power” knows that in experiment 1, it should stay out of the generator so that it is not dissipated in the source resistor but in experiment 2, it should enter the circulator so that it can be dissipated in the circulator load resistor. Can you explain how the “reflected power” “knows”? Reflected "power" doesn't know anything. Reflected voltages and currents simply respond to the known laws of physics involving wave reflection and wave superposition, and obey the conservation of energy principle. In 1, it doesn't stay out of the generator. It is redistributed from the generator back toward the load by superposition associated with *destructive interference) which happens when two superposed coherent waves are between 90 deg and 180 deg out of phase. The entire experiment is set up in a perfect 50 ohm environment so power = V^2/50. What you guys are missing is that step which carries the energy along with the voltage, i.e. the voltage (and current) require a certain energy level which is being ignored. In 2, the source wave never encounters the reflected wave so there is no superposition or interference at the source resistor. You can add a couple of more configurations. Number 3 could be when the interference between the source wave and the reflected wave occurs and they are less than 90 deg out of phase. More power than the average reflected power plus average load power will be dissipated in the source resistor because of superposition associated with *constructive interference*. Number 4 could be the special case when the source wave and the reflected wave are 90 degrees out of phase. This is the condition of zero interference (neither destructive nor constructive) and 100% of the average reflected power is dissipated in the source resistor. This is the easiest case to understand because there are no reflections and no interference. -- 73, Cecil, w5dxp.com Dear Cecil: I understand you proposition about energy redistribution on reflections and constructive/destructive interference phenomena. It is easy to me understand tridimensional waves examples interfering on a surface, rendering strips of nulls and maximuns. In a TL I also understand interference must render -for example- a destructive composition towar one direction and a constructive towards opposite direction. Full destructive interference in one directions implies full constructive on the opposite one, rendering a unidirectional energy flow towards the constructive direction, and zero flow toward the other. However I can not visualize a simple mechanism to generate such system in a TL. Tridimensional examples are easily because RF or light source can be set very close to each other as in physics traditional ligth examples and render the interference pattern, but in a TL they need be in the same place: I managed to explain my idea? Give me a hand. (It is a pitty newsgroups do not have image capabilities) I am studying your paper published in World Radio Oct 2005. Have you a demonstration of the equation cited fro "Optics" book = Pfor=Pi +P2+2[sqrt(P1P2)]cos(theta)? 73 - Miguel - LU6ETJ |
what happens to reflected energy ?
On Jun 23, 10:06*pm, lu6etj wrote:
On 22 jun, 11:45, Cecil Moore wrote: On Jun 22, 5:49*am, Keith Dysart wrote: What puzzles me then, is how the “reflected power” knows that in experiment 1, it should stay out of the generator so that it is not dissipated in the source resistor but in experiment 2, it should enter the circulator so that it can be dissipated in the circulator load resistor. Can you explain how the “reflected power” “knows”? Reflected "power" doesn't know anything. Reflected voltages and currents simply respond to the known laws of physics involving wave reflection and wave superposition, and obey the conservation of energy principle. In 1, it doesn't stay out of the generator. It is redistributed from the generator back toward the load by superposition associated with *destructive interference) which happens when two superposed coherent waves are between 90 deg and 180 deg out of phase. The entire experiment is set up in a perfect 50 ohm environment so power = V^2/50. What you guys are missing is that step which carries the energy along with the voltage, i.e. the voltage (and current) require a certain energy level which is being ignored. In 2, the source wave never encounters the reflected wave so there is no superposition or interference at the source resistor. You can add a couple of more configurations. Number 3 could be when the interference between the source wave and the reflected wave occurs and they are less than 90 deg out of phase. More power than the average reflected power plus average load power will be dissipated in the source resistor because of superposition associated with *constructive interference*. Number 4 could be the special case when the source wave and the reflected wave are 90 degrees out of phase. This is the condition of zero interference (neither destructive nor constructive) and 100% of the average reflected power is dissipated in the source resistor. This is the easiest case to understand because there are no reflections and no interference. -- 73, Cecil, w5dxp.com Dear Cecil: I understand you proposition about energy redistribution on reflections and constructive/destructive interference phenomena. It is easy to me understand tridimensional waves examples interfering on a surface, rendering strips of nulls and maximuns. In a TL I also understand interference must render -for example- a destructive composition towar one direction and a constructive towards opposite direction. Full destructive interference in one directions implies full constructive on the opposite one, rendering a unidirectional energy flow towards the constructive direction, and zero flow toward the other. However I can not visualize a simple mechanism to generate such system in a TL. Tridimensional examples are easily *because RF or light source can be set very close to each other as in physics traditional ligth examples and render the interference pattern, but in a TL they need be in the same place: I managed to explain my idea? Give me a hand. (It is a pitty newsgroups do not have image capabilities) I am studying your paper published in World Radio Oct 2005. Have you a demonstration of the equation cited fro "Optics" book = Pfor=Pi +P2+2[sqrt(P1P2)]cos(theta)? This equation is problematic. Firstly, it mixes power with voltage since theta is the angle between the voltage waveforms. As K1TTT said, "why not just do the whole thing with voltages?" This mixing is bad form and clearly demonstrates the incompleteness of the power based analysis. The second is that there are two solutions depending on whether the positive or negative root is used? Why is one discarded? Is this numerology at work? Thirdly, it only produces the correct answer for average energy flows. If the instantaneous energy flows are examined, the results using this equation do not align with observations. ....Keith |
what happens to reflected energy ?
On Jun 23, 9:06*pm, lu6etj wrote:
However I can not visualize a simple mechanism to generate such system in a TL. I have given the equations for what happens at an impedance discontinuity in a transmission line. The s-parameter equations are the same equations in a different format. Simply visualize the voltage phasors resulting from reflections-from and transmissions-through the impedance discontinuity. Let's start with voltages instead of power. source------Z01=50 ohms------+------Z02=300 ohms--------load measured Vfor1 = 50v, Vref1 = 0v calculated rho1 = (300-50)/(300+50) = 0.7143 Vfor1*rho1 = 35.7v at zero degrees = portion of Vfor1 reflected by the impedance discontinuity at '+'. Vref2*tau2 = 35.7v at 180 degrees = portion of Vref2 transmitted back through the impedance discontinuity at '+'. Vref1 = Vfor*rho1 + Vref2*tau2 = 0, measured Vref1. This is the same as the s-parameter equation (1). b1 = s11*a1 + s12*a2, all phasor math These are the two voltage components that superpose to zero volts. Both of those wavefronts are phasors with phase angles referenced to the Vfor1 phase angle. Now let's look at the power in the component phasor wavefronts. (Vfor1)^2/Z01 = 50^2/50 = 50w, power in the Vfor1 forward wave (Vref1)^2/Z01 = 0^2/50 = 0w, power in the Vref1 reflected wave (Vfor1*rho1)^2/Z01 = 35.7^2/50 = 25.49w (Vref2*tau2)^2/Z01 = 35.7^2/50 = 25.49w Pref1 = 25.49w + 25.49w + 2*SQRT(25.49*25.49)cos(180) The corresponding s-parameter power equation is: b1^2 = (s11*a1 + s12*a2)^2 b1^2 = (s11*a1)^2 + (s12*a2)^2 + 2(s11*a1)(s12*a2) The two wavefronts that cancel toward the source contain energy before they superpose to zero. Where does that energy go? Superposing to zero indicates total destructive interference toward the source. The conservation of energy principle says that energy cannot be destroyed so it must appear as an equal magnitude of constructive interference in the only other direction possible, i.e. toward the load. The above equations deal only with the destructive interference toward the source. There is a second complimentary set of voltage/energy equations that deal with constructive interference toward the load. I am studying your paper published in World Radio Oct 2005. Have you a demonstration of the equation cited fro "Optics" book = Pfor=Pi +P2+2[sqrt(P1P2)]cos(theta)? See above. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 24, 5:17*am, Keith Dysart wrote:
I assume you will claim that there is now “constructive interference” rather than the previous “destructive interference”, but the line conditions are the same. How does the “reflected power” know if it should construct or destroy? The phase is the same. That's easy to answer. The Norton equivalent is a current source so currents should be used in the calculations. The phase angles between the two current components are 180 degrees different from the phase angles between the two voltage components. If the interference between voltages is constructive, the interference between currents will be destructive. Hint: the reflected current phasor is 180 degrees out of phase with the reflected voltage phasor because of the direction of travel of the reflected wave. As a result of directional convention, the power in the reflected wave is negative. So destructive interference for forward/reverse voltages is constructive interference for forward/reverse currents and vice versa. An SWR voltage maximum (constructive voltage interference) is an SWR current minimum (destructive current interference) and an SWR voltage minimum (destructive current interference) is an SWR current maximum (constructive voltage interference). -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On 24 jun, 10:32, Cecil Moore wrote:
On Jun 23, 9:06*pm, lu6etj wrote: However I can not visualize a simple mechanism to generate such system in a TL. I have given the equations for what happens at an impedance discontinuity in a transmission line. The s-parameter equations are the same equations in a different format. Simply visualize the voltage phasors resulting from reflections-from and transmissions-through the impedance discontinuity. Let's start with voltages instead of power. source------Z01=50 ohms------+------Z02=300 ohms--------load measured Vfor1 = 50v, Vref1 = 0v calculated rho1 = (300-50)/(300+50) = 0.7143 Vfor1*rho1 = 35.7v at zero degrees = portion of Vfor1 reflected by the impedance discontinuity at '+'. Vref2*tau2 = 35.7v at 180 degrees = portion of Vref2 transmitted back through the impedance discontinuity at '+'. Vref1 = Vfor*rho1 + Vref2*tau2 = 0, measured Vref1. This is the same as the s-parameter equation (1). b1 = s11*a1 + s12*a2, all phasor math These are the two voltage components that superpose to zero volts. Both of those wavefronts are phasors with phase angles referenced to the Vfor1 phase angle. Now let's look at the power in the component phasor wavefronts. (Vfor1)^2/Z01 = 50^2/50 = 50w, power in the Vfor1 forward wave (Vref1)^2/Z01 = 0^2/50 = 0w, power in the Vref1 reflected wave (Vfor1*rho1)^2/Z01 = 35.7^2/50 = 25.49w (Vref2*tau2)^2/Z01 = 35.7^2/50 = 25.49w Pref1 = 25.49w + 25.49w + 2*SQRT(25.49*25.49)cos(180) The corresponding s-parameter power equation is: b1^2 = (s11*a1 + s12*a2)^2 b1^2 = (s11*a1)^2 + (s12*a2)^2 + 2(s11*a1)(s12*a2) The two wavefronts that cancel toward the source contain energy before they superpose to zero. Where does that energy go? Superposing to zero indicates total destructive interference toward the source. The conservation of energy principle says that energy cannot be destroyed so it must appear as an equal magnitude of constructive interference in the only other direction possible, i.e. toward the load. The above equations deal only with the destructive interference toward the source. There is a second complimentary set of voltage/energy equations that deal with constructive interference toward the load. I am studying your paper published in World Radio Oct 2005. Have you a demonstration of the equation cited fro "Optics" book = Pfor=Pi +P2+2[sqrt(P1P2)]cos(theta)? See above. -- 73, Cecil, w5dxp.com Oh, I'm so sorry Cecil, I should have written "However I can not visualize a simple PHYSICAL mechanism/example to generate such system in a TL". Anyway, your additional info it is very usefu to me. Thanks. Miguel |
what happens to reflected energy ?
On Jun 24, 5:27*am, Keith Dysart wrote:
This equation is problematic. Firstly, it mixes power with voltage since theta is the angle between the voltage waveforms. You apparently don't understand what happens when one takes the dot product of two voltage phasors (and divides by Z0). The result is watts but the math involves the cosine of the angle between the two voltage phasors. All competent EEs should already know that. As K1TTT said, "why not just do the whole thing with voltages?" Because the title of this thread is: "What happens to reflected energy?", not what happens to reflected voltages? Doing the whole thing with voltages allows the obfuscation of interference to be swept under the rug. Because some of you guys don't recognize interference when it is staring you in face? I was taught to recognize interference between voltage phasors at Texas A&M in the 1950s. What happened to you guys? Here's a short lesson about dot products of voltage phasors and the resulting interference between the two voltages. Vtot = V1*V2 Vtot^2 = (V1*V2)^2 Vtot^2/Z0 watts = (V1*V2)^2/Z0 watts There will be the two obvious power terms, V1^2/Z0 and V2^2/Z0, representing the powers in the individual waves before superposition. There will be a third, additional interference term whose dimension is watts that *requires the dot product* between the two phasor voltages. Therefore, your objection is apparently just based on ignorance of the dot product of two voltage phasors. This mixing is bad form and clearly demonstrates the incompleteness of the power based analysis. Good grief! This "bad form" has been honored in the field of optical physics for at least a century. It was taught in EE courses 60 years ago. I don't know what has happened in the meantime. Walter Maxwell explains interference in section 4.3 in "Reflections" and obviously understands the role of interference in the redistribution of energy. The second is that there are two solutions depending on whether the positive or negative root is used? Why is one discarded? Is this numerology at work? Negative power is just a convention for "negative" direction of energy flow. All EEs are taught in our engineering courses to ignore the imiginary root when calculating resistance, energy, or power. For instance, the Z0 for the 1/4WL matching section between R1 and R2 needs to be SQRT(R1*R2). When you perform that math function, do you really go on a world-wide search demanding a transmssion line with a negative Z0? Please get real. Thirdly, it only produces the correct answer for average energy flows. If the instantaneous energy flows are examined, the results using this equation do not align with observations. You forgot to add that instantaneous energy is as useless as tits on a boar hog, or as Hecht said, putting it mildly: "of limited utility". It appears to me that instantaneous energy is just a mathematical artifact inside a process requiring integration in order to bear any resemblence to reality. Omit the integration and the process loses touch with reality. Instantaneous energy has zero area under the curve until the intergration process has been performed. A zero area represents zero energy. Otherwise, when you integrate from zero to infinity, the result would be infinite energy. Do you have any kind of reference for your treatment of instantaneous power? -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 24, 9:20*am, lu6etj wrote:
Oh, I'm so sorry Cecil, I should have written "However I can not visualize a simple PHYSICAL mechanism/example to generate such system in a TL". Anyway, your additional info it is very useful to me. Thanks. The physical mechanism is the Z01==Z02 impedance discontinuity with its associated reflection coefficient, rho. We can see that reflection on a TDR so it is indeed a PHYSICAL mechanism. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 24, 9:39*am, Cecil Moore wrote:
Vtot^2/Z0 watts = (V1*V2)^2/Z0 watts There will be the two obvious power terms, V1^2/Z0 and V2^2/Z0, Sorry, I lost my train-of-thought here and changed horses in mid- stream. All I was trying to illustrate is that the dot product of two voltage phasors divided by Z0 has the dimensions of watts and contains the cos(A) term. Again, my apology for wandering off the logical path. -- 73, Cecil, w5dxp.com |
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